Let's solve the given equation:

\[
\sin x + \frac{2pq \cos x}{p^2 - q^2} = 1
\]

To find \(\tan x\), let's denote \(\tan x = t\). We will use the identity \(\sin x = \frac{t}{\sqrt{1+t^2}}\) and \(\cos x = \frac{1}{\sqrt{1+t^2}}\).

### Step 1: Substitute \(\sin x\) and \(\cos x\) in terms of \(t\).
Substitute these into the given equation:

\[
\frac{t}{\sqrt{1+t^2}} + \frac{2pq \cdot \frac{1}{\sqrt{1+t^2}}}{p^2 - q^2} = 1
\]

Simplifying:

\[
\frac{t + \frac{2pq}{p^2 - q^2}}{\sqrt{1+t^2}} = 1
\]

### Step 2: Multiply both sides by \(\sqrt{1+t^2}\).

\[
t + \frac{2pq}{p^2 - q^2} = \sqrt{1+t^2}
\]

Square both sides:

\[
\left(t + \frac{2pq}{p^2 - q^2}\right)^2 = 1 + t^2
\]

### Step 3: Expand and simplify.

\[
t^2 + \frac{4pqt}{p^2 - q^2} + \frac{4p^2q^2}{(p^2 - q^2)^2} = 1 + t^2
\]

Cancel \(t^2\) on both sides:

\[
\frac{4pqt}{p^2 - q^2} + \frac{4p^2q^2}{(p^2 - q^2)^2} = 1
\]

### Step 4: Further simplify and solve for \(t\).

\[
\frac{4pq(p^2 - q^2)t + 4p^2q^2}{(p^2 - q^2)^2} = 1
\]

Multiply by \((p^2 - q^2)^2\):

\[
4pq(p^2 - q^2)t + 4p^2q^2 = (p^2 - q^2)^2
\]

\[
4pq(p^2 - q^2)t = (p^2 - q^2)^2 - 4p^2q^2
\]

\[
t = \frac{(p^2 - q^2)^2 - 4p^2q^2}{4pq(p^2 - q^2)}
\]

Simplifying:

\[
t = \frac{(p^2 + q^2)(p^2 - q^2)}{4pq(p^2 - q^2)}
\]

Finally, the solution for \(\tan x\) is:

\[
t = \frac{p^2 + q^2}{4pq}
\]

So, \(\tan x = \frac{p^2 + q^2}{4pq}\).

Would you like further details or have any questions?

### Related Questions:
1. How do you find \(\sin x\) and \(\cos x\) using the value of \(\tan x\)?
2. What are the trigonometric identities used in solving such equations?
3. How can you solve similar equations with different trigonometric functions?
4. What are the applications of \(\tan x\) in real-world problems?
5. How do trigonometric equations relate to algebraic identities?

**Tip:** Understanding the relationships between different trigonometric functions is crucial for simplifying and solving complex equations.

Sinx + 2pqcosx / (p^2 - q^2) = 1, find tanx

Learn how to solve the equation Sinx + 2pqcosx / (p^2 - q^2) = 1 and find the value of tanx. This solution includes the use of trigonometric identities such as the Pythagorean identity, and involves algebraic manipulation. Suitable for students in Grades 11-12.

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